Sara Solomon

2022-10-08

I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example:

$3{y}^{2}=5{x}^{3}$

I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:

$\frac{d}{dx}(3{y}^{2})=\frac{d}{dx}(5{x}^{3})$

$6y\frac{d}{dx}(y)=15{x}^{2}\frac{d}{dx}(x)$

$6y\frac{dy}{dx}=15{x}^{2}\frac{dx}{dx}$

$6y\frac{dy}{dx}=15{x}^{2}$

$\frac{dy}{dx}=\frac{15{x}^{2}}{6y}$

Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

$3{y}^{2}=5{x}^{3}$

I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:

$\frac{d}{dx}(3{y}^{2})=\frac{d}{dx}(5{x}^{3})$

$6y\frac{d}{dx}(y)=15{x}^{2}\frac{d}{dx}(x)$

$6y\frac{dy}{dx}=15{x}^{2}\frac{dx}{dx}$

$6y\frac{dy}{dx}=15{x}^{2}$

$\frac{dy}{dx}=\frac{15{x}^{2}}{6y}$

Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

Shane Middleton

Beginner2022-10-09Added 7 answers

$\frac{\mathrm{d}y}{\mathrm{d}x}$ may equal 1 in special cases, but it is not generally so. The very point of implicit derivation is that you don't know what it is.

So, the Chain Rule is used to separate the derivative of a function of $y$ wrt $x$ into two differential terms, one you can resolve, and one unknown.

To be clear, here's the process with the stage added and highlighted. Does this help?

$\begin{array}{rl}3{y}^{2}& =5{x}^{3}\\ \frac{\mathrm{d}(3{y}^{2})}{\mathrm{d}x}& =\frac{\mathrm{d}(5{x}^{3})}{\mathrm{d}x}& \text{Take the derivative w.r.t.}x\\ {\frac{\mathrm{d}(3{y}^{2})}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}}& =\frac{\mathrm{d}(5{x}^{3})}{\mathrm{d}x}& \text{Apply the Chain Rule to the L.H.S.}\\ 6y\frac{\mathrm{d}y}{\mathrm{d}x}& =15{x}^{2}& \text{Evaluate the polynomial derivatives}\frac{\mathrm{d}(c\phantom{\rule{thinmathspace}{0ex}}{z}^{n})}{\mathrm{d}z}=c\phantom{\rule{thinmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}{z}^{n-1}\\ & & \text{where}n\in \mathbb{N},\text{and}c\text{is constant}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{5{x}^{2}}{2y}& \text{Use arthimetic rearrangement}\end{array}$

Alternatively, we might use explicit differentiation, as follows:

$\begin{array}{rl}3{y}^{2}& =5{x}^{3}\\ y& ={x}^{3/2}\phantom{\rule{thinmathspace}{0ex}}\sqrt{\frac{5}{3}\phantom{\rule{thickmathspace}{0ex}}}& \star \\ \frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{\mathrm{d}{x}^{3/2}}{\mathrm{d}x}\phantom{\rule{thickmathspace}{0ex}}\sqrt{\frac{5}{3}\phantom{\rule{thickmathspace}{0ex}}}\\ \text{}& =\frac{3{x}^{1/2}\sqrt{5}}{2\sqrt{3}}\\ \text{}& =\frac{3{x}^{1/2}\sqrt{5}}{2\sqrt{3}}\times \frac{{x}^{3/2}\sqrt{5}}{y\sqrt{3}}& \text{re:}\star \\ \text{}& =\frac{5{x}^{2}}{2y}\end{array}$

So, the Chain Rule is used to separate the derivative of a function of $y$ wrt $x$ into two differential terms, one you can resolve, and one unknown.

To be clear, here's the process with the stage added and highlighted. Does this help?

$\begin{array}{rl}3{y}^{2}& =5{x}^{3}\\ \frac{\mathrm{d}(3{y}^{2})}{\mathrm{d}x}& =\frac{\mathrm{d}(5{x}^{3})}{\mathrm{d}x}& \text{Take the derivative w.r.t.}x\\ {\frac{\mathrm{d}(3{y}^{2})}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}}& =\frac{\mathrm{d}(5{x}^{3})}{\mathrm{d}x}& \text{Apply the Chain Rule to the L.H.S.}\\ 6y\frac{\mathrm{d}y}{\mathrm{d}x}& =15{x}^{2}& \text{Evaluate the polynomial derivatives}\frac{\mathrm{d}(c\phantom{\rule{thinmathspace}{0ex}}{z}^{n})}{\mathrm{d}z}=c\phantom{\rule{thinmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}{z}^{n-1}\\ & & \text{where}n\in \mathbb{N},\text{and}c\text{is constant}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{5{x}^{2}}{2y}& \text{Use arthimetic rearrangement}\end{array}$

Alternatively, we might use explicit differentiation, as follows:

$\begin{array}{rl}3{y}^{2}& =5{x}^{3}\\ y& ={x}^{3/2}\phantom{\rule{thinmathspace}{0ex}}\sqrt{\frac{5}{3}\phantom{\rule{thickmathspace}{0ex}}}& \star \\ \frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{\mathrm{d}{x}^{3/2}}{\mathrm{d}x}\phantom{\rule{thickmathspace}{0ex}}\sqrt{\frac{5}{3}\phantom{\rule{thickmathspace}{0ex}}}\\ \text{}& =\frac{3{x}^{1/2}\sqrt{5}}{2\sqrt{3}}\\ \text{}& =\frac{3{x}^{1/2}\sqrt{5}}{2\sqrt{3}}\times \frac{{x}^{3/2}\sqrt{5}}{y\sqrt{3}}& \text{re:}\star \\ \text{}& =\frac{5{x}^{2}}{2y}\end{array}$

Sanai Ball

Beginner2022-10-10Added 1 answers

Change $y$ to $y(x)$ and everything should be clear. $y$ is not a constant, but a function of $x$.

$\frac{dy}{dx}=\frac{d}{dx}y(x)={y}^{\prime}(x)$

$\frac{dy}{dx}=\frac{d}{dx}y(x)={y}^{\prime}(x)$

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